How Does Time Dilation Affect Supersonic Jet Pilots?

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SUMMARY

Time dilation significantly affects supersonic jet pilots, with calculations showing that during a year (3.16 * 10^7 seconds) on an observer's clock, only 31600000.0001422 seconds elapse on the pilot's clock. This results in a negligible time difference of approximately 0.0001422 seconds, or 2.37e-5 minutes, indicating that the pilot experiences less time due to relativistic effects. The proper time, defined as the time measured by an inertial clock at both events, confirms that the pilot's clock runs slower relative to the observer's clock.

PREREQUISITES
  • Understanding of special relativity concepts, particularly time dilation.
  • Familiarity with the equation for time dilation: delta(t) = delta(t') / sqrt(1 - v^2/c^2).
  • Basic knowledge of inertial reference frames and proper time.
  • Ability to perform calculations involving scientific notation and square roots.
NEXT STEPS
  • Study the implications of time dilation in high-speed travel scenarios.
  • Learn about the Lorentz transformation and its applications in physics.
  • Explore the concept of proper time in more detail through advanced relativity texts.
  • Investigate real-world applications of time dilation in GPS technology and satellite communications.
USEFUL FOR

Physics students, aerospace engineers, and anyone interested in the effects of relativistic speeds on time perception and measurement.

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Homework Statement


Supersonic jets achieve maximum speeds of about (3 *10^6)*c.

a)During a time of 1 y =3.16*10^7 seconds on your clock, how much time would elapse on the pilot's clock?

b)How many minutes are lost by the pilots clock in 1 year of your time.

Homework Equations



delta(t)=delta(t')/(sqrt(1-v^2/c^2)^.5), where delta(t') is the proper time interval

the proper time would be the time elapse(delta(t')) on my clock and the time dilated (delta (t)) would be the pilot's clock.

The Attempt at a Solution



to solved the first problem, I would have to apply the equations I mentioned and 2 and solved delta(t). delta(t)= (3.16e7)/(sqrt(1-(3e-6)^2)).

delta(t)=31600000.0001422000000095985 seconds . So I say the time dilated on the pilot's clock is very negligble.

The second part of the question just asked the time difference between my clock time and the pilot clock's time and that results to be. delta(t)=.0001422000000095985 seconds =2.37e-5 minutes .
 
Last edited:
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You have one mistake here, other than that your work seems great, assuming you have not made an error while crunching the numbers. You have the wrong time as the proper time. The proper time is time the pilot experiences. This is because the proper time is the time given by an inertial clock present and at both events. Your clock is always with you, but the pilots clock is actually there at the events you are measuring (the beginning and end of the plane's year long trip).

(I like this description of proper time, but I found it on Wikipedia I think, so I am a little suspicious that it may not be a complete description of proper time. The definition seems to be fine to me, but if someone finds that it is not complete, please add on to it.)

Remember that the pilot should experience less time relative to you since he is moving relative to you, yet your answer says the pilot experiences MORE time. This is what gave the mistake away.

Other than that, your solution seems correct. Nice Job!
 
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